61. If 1 man or 2 women or 3 boys can complete a piece of work in 88 days, then 1 man, 1 woman and 1 boy together will complete it in

1. 1. 36 days

   
   
   

   2. 42 days

   
   
   

   3. 48 days

   
   
   

   4. 54 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   1 man = 2 women ≡ 3 boys
   1 man + 1 woman + 1 boy
   

   =		3 +	3	+ 1		boys =	11	boys
   			20				20

   
   ∴ M₁D₁ = M₂D₂
   

   ⇒ 3 × 88 =	11	× D2
   	2

   
   

   ⇒ D2 =	2 × 3 × 88	= 48 days
   	11

   
   Aliter :
   Using Rule 13,
   Here, A = 1, B = 2, C = 3, a = 88 A₁ = 1, B₁ = 1, C₁ = 1
   

   Time taken =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	88
   	1	+	1	+	1
   	1		2		3

   
   

   =	88 × 6	= 48 days
   	6 + 3 + 2

   Correct Option: C

   1 man = 2 women ≡ 3 boys
   1 man + 1 woman + 1 boy
   

   =		3 +	3	+ 1		boys =	11	boys
   			20				20

   
   ∴ M₁D₁ = M₂D₂
   

   ⇒ 3 × 88 =	11	× D2
   	2

   
   

   ⇒ D2 =	2 × 3 × 88	= 48 days
   	11

   
   Aliter :
   Using Rule 13,
   Here, A = 1, B = 2, C = 3, a = 88 A₁ = 1, B₁ = 1, C₁ = 1
   

   Time taken =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	88
   	1	+	1	+	1
   	1		2		3

   
   

   =	88 × 6	= 48 days
   	6 + 3 + 2

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62. A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days ?

1. 1. 8

   
   
   

   2. 12

   
   
   

   3. 4

   
   
   

   4. 6

   
   
   

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1. View Hint View Answer Discuss in Forum

   Part of work done by 2 men and 2 women in 2 days.
   

   = 2		2	+	8
   		20		30

   
   

   = 2		1	+	8		= 2		3 + 8
   		10		30				30

   
   

   =	22	=	11
   	30		15

   
   

   Remaining work = 1 -	11	=	4
   	15		15

   
   Work done by 1 boy in 2 days
   

   =	2	=	1
   	60		30

   
   ∴ Number of boys required to assist
   

   =	4	× 30 = 8
   	15

   
   Aliter : Using Rule 14,
   Here, A = 1, B = 1, C = 1
   a = 20, b = 30, c = 60
   A₁ = 2, B₁ = 8,
   Required time
   

   =	1
   	A1	+	B1	+	C1
   	A × a		B × b		C × c

   
   

   2 =	1
   	2	+	8	+	x
   	1 × 20		1 × 30		1 × 60

   
   

   2 =	10
   	2	+	8	+	x
   	2		3		6

   
   

   2 =	10
   	6 + 16 + x
   	6

   
   22 + x = 30 x = 8
   ∴ Number of boys = 8

   Correct Option: A

   Part of work done by 2 men and 2 women in 2 days.
   

   = 2		2	+	8
   		20		30

   
   

   = 2		1	+	8		= 2		3 + 8
   		10		30				30

   
   

   =	22	=	11
   	30		15

   
   

   Remaining work = 1 -	11	=	4
   	15		15

   
   Work done by 1 boy in 2 days
   

   =	2	=	1
   	60		30

   
   ∴ Number of boys required to assist
   

   =	4	× 30 = 8
   	15

   
   Aliter : Using Rule 14,
   Here, A = 1, B = 1, C = 1
   a = 20, b = 30, c = 60
   A₁ = 2, B₁ = 8,
   Required time
   

   =	1
   	A1	+	B1	+	C1
   	A × a		B × b		C × c

   
   

   2 =	1
   	2	+	8	+	x
   	1 × 20		1 × 30		1 × 60

   
   

   2 =	10
   	2	+	8	+	x
   	2		3		6

   
   

   2 =	10
   	6 + 16 + x
   	6

   
   22 + x = 30 x = 8
   ∴ Number of boys = 8

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63. 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it ?

1. 1. 50 days

   
   
   

   2. 45 days

   
   
   

   3. 40 days

   
   
   

   4. 35 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let 1 man’s 1 day’s work = x and
   1 woman’s 1 day’s work = y
   

   Then, 4x + 6y =	1	and
   	8

   
   

   3x + 7y =	1
   	10

   
   From both equations,
   

   we get y =	1
   	400

   
   ∴ 10 women’s 1 day’s work
   

   =	10	=	1
   	400		40

   
   ∴ 10 women will finish the work in 40 days.
   Aliter : Using Rule 11,
   A₁ = 4, B₁ = 6, D₁ = 8
   A₂ = 3, B₂ = 7, D₂ = 10
   A₃ = 0, B₃ = 10
   Required time
   

   =	D1D2(A1B2 - A2B1)
   	D1(A1B3 - A3B1) - D2(A2B3 - A3B2)

   
   

   =	8 × 10(4 × 7 - 3 × 6)
   	8(4 × 10 - 0 × 6) - 10(3 × 10 - 0 × 7)

   
   

   =	80 × 10	= 40 days
   	20

   Correct Option: C

   Let 1 man’s 1 day’s work = x and
   1 woman’s 1 day’s work = y
   

   Then, 4x + 6y =	1	and
   	8

   
   

   3x + 7y =	1
   	10

   
   From both equations,
   

   we get y =	1
   	400

   
   ∴ 10 women’s 1 day’s work
   

   =	10	=	1
   	400		40

   
   ∴ 10 women will finish the work in 40 days.
   Aliter : Using Rule 11,
   A₁ = 4, B₁ = 6, D₁ = 8
   A₂ = 3, B₂ = 7, D₂ = 10
   A₃ = 0, B₃ = 10
   Required time
   

   =	D1D2(A1B2 - A2B1)
   	D1(A1B3 - A3B1) - D2(A2B3 - A3B2)

   
   

   =	8 × 10(4 × 7 - 3 × 6)
   	8(4 × 10 - 0 × 6) - 10(3 × 10 - 0 × 7)

   
   

   =	80 × 10	= 40 days
   	20

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64. One man and one woman together can complete a piece of work in 8 days. A man alone can complete the work in 10 days. In how many days can one woman alone complete the work ?

1. 1. 	140	days
      	9

   
   
   

   2. 30 days

   
   
   

   3. 40 days

   
   
   

   4. 42 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   Work done by 1 woman in 1 day
   

   =	1	-	1	=	5 - 4	=	1
   	8		10		40		40

   
   ∴ One woman will complete the work in 40 days.
   Aliter : Using Rule 4,
   Here, x = 10 and y = 8 Woman can do work in
   

   =			xy		days
   			x - y

   
   

   =			10 × 8		= 40 days
   			10 - 8

   Correct Option: C

   Work done by 1 woman in 1 day
   

   =	1	-	1	=	5 - 4	=	1
   	8		10		40		40

   
   ∴ One woman will complete the work in 40 days.
   Aliter : Using Rule 4,
   Here, x = 10 and y = 8 Woman can do work in
   

   =			xy		days
   			x - y

   
   

   =			10 × 8		= 40 days
   			10 - 8

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65. Three men can complete a piece of work in 6 days. Two days after they started the work, 3 more men joined them. How many days will they take to complete the remaining work ?

1. 1. 1 days

   
   
   

   2. 2 days

   
   
   

   3. 3 days

   
   
   

   4. 4 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   Work done in two days =	1	× 2
   	6

   
   

   =	1	, remaining work =	2
   	3		3

   
   

   ⇒	M1D1	=	M2D2
   	W1		W2

   
   

   ⇒	3 × 2	=	6 × D2
   	1		2
   	3		3

   
   

   ⇒ D2 =	3 × 2 × 2	= 2 days
   	6

   
   Aliter : Using Rule 10,
   Here, A = 3, a = 6
   b = 2, B = 3
   

   Required time =	A(a - b)
   	A + B

   
   

   =	3(6 - 2)
   	3 + 3

   
   

   =	3 × 4	= 2 days
   	6

   Correct Option: B

   Work done in two days =	1	× 2
   	6

   
   

   =	1	, remaining work =	2
   	3		3

   
   

   ⇒	M1D1	=	M2D2
   	W1		W2

   
   

   ⇒	3 × 2	=	6 × D2
   	1		2
   	3		3

   
   

   ⇒ D2 =	3 × 2 × 2	= 2 days
   	6

   
   Aliter : Using Rule 10,
   Here, A = 3, a = 6
   b = 2, B = 3
   

   Required time =	A(a - b)
   	A + B

   
   

   =	3(6 - 2)
   	3 + 3

   
   

   =	3 × 4	= 2 days
   	6

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